I have no doubt that the identity is true, but I am not able to prove it. Can anyone help?

It is easy to prove by Taylor expansion that the left-hand-side of the identity can equivalently be written as $\sum_{n=1}^{\infty}\ln(1+x^n)$, which is the logarithm of the q-Pochhammer symbol $(-x,x)_{\infty}$, so an alternative way to pose my question is to ask for a proof of the series expansion

@Gjergji Zaimi Thanks for the help. I could not derive any of the 2 equalities that you stated in the beginning. Can you kindly give some proof or reference for them? Especially in the second one I don't see any occurrence of the summing variable k in the summand. Can you kindly add some more lines of explanation!
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AnirbitNov 19 '12 at 23:56

I may try to add some more explanations a bit later, but if you expand the fractions $\frac{1}{1-x^n}=1+x^n+x^{2n}+\cdots$, it shouldn't be very hard to see the equations that I wrote.
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Gjergji ZaimiNov 20 '12 at 1:52

I would make a mere comment since Gjergji has already answered, but I am not allowed to make comments.

... so an alternative way to pose my question is to ask for a proof of the series expansion
$\ln(-x,x)_{\infty}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^{n}}{1-x^{2n}}$.

This is a corollary of Euler's theorem that the number of partitions of $n$ into distinct parts is equal to the number of partitions of $n$ into odd parts. In terms of generating functions, Euler's theorem is just
$(-x,x)_{\infty}=\frac{1}{(x,x^2)_\infty}$, which can be easily proved by replacing the
term $(1+x^i)$ in $(-x,x)_\infty$ by $\frac{1-x^{2i}}{1-x^i}$ and cancelling all the terms in
the numerator against the corresponding terms in the denominator. By Euler's theorem,
$\ln \left( (-x,x)_{\infty}\right) =\ln\left( \frac{1}{(x,x^2)_\infty}\right)
=\sum_{i=1}^\infty \ln \left( \frac{1}{1-x^{2i-1}} \right)
=\sum_{i=1}^\infty \sum_{n=1}^\infty \frac 1 n x^{n (2i-1)}$

@Garth Payne So in the light of your answer does this function $\prod _{n=1} ^{\infty} (1+x^n)$ have any modular properties? Also can you give a more specific reference to this Euler's theorem that you refer to whose special this is you say.
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AnirbitNov 21 '12 at 3:20

1

@anirbit. Euler's theorem is in George Andrew's book The Theory of Partitions (p. 5). The proof I refer to above is given there in more detail. In case you don't have access to the book, someone has dupiicated the proof here: math.stackexchange.com/questions/54961/… .
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Garth PayneNov 27 '12 at 15:22